package leetcode.dynamic;

/**
 * 给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
 * <p>
 * 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
 * <p>
 * <p>
 * <p>
 * 示例 1：
 * <p>
 * <p>
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
 * 输出：true
 * 示例 2：
 * <p>
 * <p>
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
 * 输出：true
 * 示例 3：
 * <p>
 * <p>
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
 * 输出：false
 * <p>
 * <p>
 * 提示：
 * <p>
 * m == board.length
 * n = board[i].length
 * 1 <= m, n <= 6
 * 1 <= word.length <= 15
 * board 和 word 仅由大小写英文字母组成
 * <p>
 * <p>
 * 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
 */
public class LeetCode79_Exist {
    public boolean exist(char[][] board, String word) {
        boolean ans = false;
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                ans = ans || f(i, j, board, word.toCharArray(), 0, new boolean[board.length][board[0].length]);
            }
        }
        return ans;
    }

    public boolean f(int i, int j, char[][] board, char[] word, int idx, boolean[][] used) {
        if (idx == word.length) {
            return true;
        }
        
        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || used[i][j]) {
            return false;
        }
        if (word[idx] == board[i][j]) {
            used[i][j] = true;
            boolean ans1 = f(i + 1, j, board, word, idx + 1, used);
            boolean ans2 = f(i - 1, j, board, word, idx + 1, used);
            boolean ans3 = f(i, j + 1, board, word, idx + 1, used);
            boolean ans4 = f(i, j - 1, board, word, idx + 1, used);
            used[i][j] = false;

            return ans1 || ans2 || ans3 || ans4;
        } else {
            return false;
        }
    }
}
